Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 24

Chapter 3, Problem 24

Once Kate’s kite reaches a height of 50 ft (above her hands), it rises no higher but drifts due east in a wind blowing 5 ft/s. How fast is the string running through Kate’s hands at the moment when she has released 120 ft of string?

Verified step by step guidance

Identify the right triangle formed by the height of the kite, the horizontal distance from Kate to the kite, and the length of the string. The height is constant at 50 ft, the horizontal distance is changing, and the string length is 120 ft at the moment of interest.

Use the Pythagorean theorem to relate the sides of the triangle: $ s^2 = x^2 + 50^2 $, where $ s $ is the length of the string and $ x $ is the horizontal distance.

Differentiate both sides of the equation with respect to time $ t $ to find the relationship between the rates of change: $ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} $.

Substitute the known values into the differentiated equation. At the moment of interest, $ s = 120 $ ft, $ \frac{dx}{dt} = 5 $ ft/s, and solve for $ \frac{ds}{dt} $, the rate at which the string is running through Kate's hands.

Calculate $ x $ using the Pythagorean theorem: $ x = \sqrt{s^2 - 50^2} $. Substitute $ x $ and the other known values into the differentiated equation to find $ \frac{ds}{dt} $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Related Rates

Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to determine how fast the string is being released as the kite drifts horizontally while maintaining a constant height. This requires applying the concept of derivatives to relate the rates of change of the kite's height, the horizontal distance, and the length of the string.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that relates the lengths of the sides of a right triangle. In this scenario, the height of the kite, the horizontal distance it drifts, and the length of the string form a right triangle. Understanding this relationship is crucial for setting up the equation that connects the variables involved in the problem.

Chain Rule

The chain rule is a formula for computing the derivative of a composite function. In the context of this problem, we will use the chain rule to differentiate the equation derived from the Pythagorean theorem with respect to time. This allows us to relate the rates of change of the string length and the horizontal distance, enabling us to find the speed at which the string is being released.

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Related Practice

Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t² + 19.6t + 24.5.

On what intervals is the speed increasing?

Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 32t + 48.

On what intervals is the speed increasing?

Textbook Question

With what velocity does the stone strike the ground?

Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 64t + 32.

On what intervals is the speed increasing?

Textbook Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is $s (t) = - 16 t^{2} + 32 t + 48$ .

d. When does the stone strike the ground?

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Textbook Question

Evaluate the derivative of the following functions.

f(x) = x² + 2x³ cot^-1 x - ln (1 + x²)